Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $22.4$ years; the standard deviation is $5.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living between $33.2$ and $38.6$ years.
Solution: $22.4$ $17$ $27.8$ $11.6$ $33.2$ $6.2$ $38.6$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $22.4$ years. We know the standard deviation is $5.4$ years, so one standard deviation below the mean is $17$ years and one standard deviation above the mean is $27.8$ years. Two standard deviations below the mean is $11.6$ years and two standard deviations above the mean is $33.2$ years. Three standard deviations below the mean is $6.2$ years and three standard deviations above the mean is $38.6$ years. We are interested in the probability of a gorilla living between $33.2$ and $38.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the gorillas will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the gorillas will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of gorillas between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular gorilla living between $33.2$ and $38.6$ years is $\color{orange}{2.35\%}$.